Let $ \{y_{1},\dots,y_{n}\} $ be an orthonormal set in a normed space $X$. Let $ Y = \operatorname{span} \{y_{1},\dots,y_{n}\} $ and $x \in X$. Check that the function $ \phi(y)=||x-y||^{2}$ has its minimum on $Y$ at $y=\sum_{i=1}^{n}\langle x, y_{i} \rangle y_{i}$ and its value is $$ ||x||^{2}-\sum_{i=1}^{n}|\langle x,y_i{} \rangle|^{2}. $$
I'm completely lost as to approach this problem and ask you for help.
This is the square of the distance function from $x$ to the finite-dimensional subspace $Y$, and by the triangular inequality its minimum is attained when $y=\pi_{_Y}(x)$, namely the orthogonal projection of $x$ to $Y$.
Since $\{y_1, \ldots, y_n\}$ is an orthonormal basis for $Y$, standard computations (that you can find in any Linear Algebra book) show that $$\pi_{_Y}(x)=\sum_{i=1}^{n}\langle x, y_{i} \rangle\, y_{i},$$ so the result follows.